description/proof of that for group with topology with continuous operations (especially, topological group), inverse of compact subset is compact and finite product of compact subsets is compact
Topics
About: group
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of group.
- The reader knows a definition of topological space.
- The reader knows a definition of continuous, topological spaces map.
- The reader knows a definition of compact subset of topological space.
- The reader admits the proposition that for any group with any topology with any continuous operations (especially, topological group) and each element, the inversion map, the multiplication-by-element-from-left-or-right map, and the conjugation-by-element map are homeomorphisms.
- The reader admits the proposition that for any continuous map between any topological spaces, the image of any compact subset of the domain is a compact subset of the codomain.
- The reader admits the proposition that the compactness of any topological subset as a subset equals the compactness as a subspace.
- The reader admits the proposition that the product of any finite number of compact topological spaces is compact.
- The reader admits the proposition that for any possibly uncountable number of indexed topological spaces or any finite number of topological spaces and their subspaces, the product of the subspaces is the subspace of the product of the base spaces.
Target Context
- The reader will have a description and a proof of the proposition that for any group with topology with continuous operations (especially, topological group), the inverse of any compact subset is compact and the finite product of any compact subsets is compact.
Orientation
There is a list of definitions discussed so far in this site.
There is a list of propositions discussed so far in this site.
Main Body
1: Structured Description
Here is the rules of Structured Description.
Entities:
\(G\): \(\in \{\text{ the groups }\}\) with any topology such that the group operations are continuous
\(K_1\): \(\in \{\text{ the compact subsets of } G\}\)
\(K_2\): \(\in \{\text{ the compact subsets of } G\}\)
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Statements:
\({K_1}^{-1} \in \{\text{ the compact subsets of } G\}\)
\(\land\)
\(K_1 K_2 \in \{\text{ the compact subsets of } G\}\)
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2: Proof
Whole Strategy: Step 1: think of the inverse map, \(f_i: G \to G, g \mapsto g^{-1}\), and see that \(f_i\) is a homeomorphism; Step 2: see that \({K_1}^{-1} = f_i (K_1)\) is compact; Step 3: think of the multiplication map, \(f_m: G \times G \to G\), and see that \(K_1 \times K_2 \subseteq G \times G\) is a compact subset; Step 4: see that \(K_1 K_2 = f_m (K_1 \times K_2)\) is compact.
Step 1:
Let us think of the inverse map, \(f_i: G \to G, g \mapsto g^{-1}\).
\(f_i\) is a homeomorphism, by the proposition that for any group with any topology with any continuous operations (especially, topological group) and each element, the inversion map, the multiplication-by-element-from-left-or-right map, and the conjugation-by-element map are homeomorphisms. Especially, \(f_i\) is continuous.
Step 2:
\({K_1}^{-1} = f_i (K_1)\).
\(f_i (K_1) \subseteq G\) is a compact subset, by the proposition that for any continuous map between any topological spaces, the image of any compact subset of the domain is a compact subset of the codomain.
So, \({K_1}^{-1} \subseteq G\) is a compact subset.
Step 3:
Let us think of the multiplication map, \(f_m: G \times G \to G\).
\(f_m\) is continuous by the supposition.
\(K_j\) is a compact subspace, by the proposition that the compactness of any topological subset as a subset equals the compactness as a subspace.
\(K_1 \times K_2\) is a compact topological space, by the proposition that the product of any finite number of compact topological spaces is compact.
\(K_1 \times K_2\) as the product of the topological subspaces is the topological subspace of \(G \times G\), by the proposition that for any possibly uncountable number of indexed topological spaces or any finite number of topological spaces and their subspaces, the product of the subspaces is the subspace of the product of the base spaces, so, \(K_1 \times K_2 \subseteq G \times G\) is a compact subspace.
\(K_1 \times K_2 \subseteq G \times G\) is a compact subset, by the proposition that the compactness of any topological subset as a subset equals the compactness as a subspace.
Step 4:
\(K_1 K_2 = f_m (K_1 \times K_2)\).
\(f_m (K_1 \times K_2) \subseteq G\) is a compact subset, by the proposition that for any continuous map between any topological spaces, the image of any compact subset of the domain is a compact subset of the codomain.
So, \(K_1 K_2 \subseteq G\) is a compact subset.
